The original post:
> Could someone help me figure out how does this circuit work? This is
> directly from Art of Electronics, but i just can't understand, how the
> gain was calculated. It was said to be 100, whereas I get 102??
I have taken the liberty of redrawing the circuit and writing a detailed
solution. First, the circuit:
100k ohm A 100k ohm
*------/\/\/\-----*-----/\/\/\----*
| | |
| | |
| |\ | |
| | \ | |
| | \ | |
100k ohm | | \ | |
Vin *----/\/\/\----*-| - \ | |
| \-------------------------* Vout
| / |
| / |
*--| + / |
| | / |
| | / |
| |/ |
\ \
/ /
\ 51k ohm \ 1k ohm
/ /
\ \
/ /
| |
| |
| |
----- -----
--- ---
- -
My solution:
By the ideal op-amp assumptions, there is no current through the 51k
ohm, so the voltage level at the inverting input terminal is zero
(ground). Then, nodal analysis at the node I have designated "A".
Summing currents away from A:
(VA - 0) VA (VA - Vout)
-------- + ---- + ----------- = 0
100k 1k 100k
By conservation, the 'source' current through the 100k between Vin and
the inverting terminal is the negate of the 'feedback' current through the
100k between the inverting terminal and A. Thus, the term
(VA - 0) Vin
-------- = - ------
100k 100k
so that VA = -Vin. Substituting this:
( -Vin - 0 ) -Vin ( -Vin - Vout )
------------ + ---- + ---------------- = 0
100k 1k 100k
Multiplying by 100k and rearranging, one winds up with Vout = -102 Vin,
for a gain magnitude of 102.
--
Paul Speicher
speicher@pollux.math.iastate.edu
Date: 28 Jun 96 04:34:27 GMT
Original Subject: Our Op-amp Problem Solved